Enter the line-to-line voltage, power factor and one known quantity (line current, active power in kW, or apparent power in kVA) to calculate the full three-phase power triangle.
Three-phase AC systems carry three quantities of power, related by the power triangle:
With V = 400 V, pf = 0.8 and line current = 10 A:
208 V (US 3-ph low)
380 V (IEC/EU)
400 V (IEC/EU)
415 V (UK)
480 V (US industrial)
VL-N = VL-L / √3
e.g. 400 V L-L ≈ 231 V L-N
e.g. 480 V L-L ≈ 277 V L-N
Resistive loads: pf = 1.0
Motors: pf 0.7 to 0.9
Fluorescent: pf ~0.85
Ideal target: pf ≥ 0.95
The power triangle shows how active power (kW), reactive power (kVAR) and apparent power (kVA) relate in an AC circuit. kVA is the hypotenuse, kW is the adjacent side, and kVAR is the opposite side. The power factor equals kW divided by kVA, which is the cosine of the angle between them.
kW (kilowatts) is real, active power that does useful work. kVA (kilovolt-amperes) is apparent power, the product of voltage and current. kVAR (kilovolt-amperes reactive) is the reactive power consumed by inductors and capacitors. The relationship is: kVA² = kW² + kVAR².
In a balanced three-phase system, the total power is the vector sum of three equal single-phase contributions displaced by 120 degrees. When you express this using the line-to-line voltage, the geometry works out to a factor of √3 (approximately 1.732). For single-phase, the factor is simply 2 (one conductor out, one return).
Line-to-line (L-L) voltage is measured between any two of the three phase conductors; line-to-neutral (L-N) voltage is measured from one phase to the neutral. They are related by VL-L = √3 × VL-N. This calculator uses line-to-line voltage because that is the value used in the three-phase power formula S = √3 × VL-L × I.
Rearrange S = √3 × V × I to get I = S / (√3 × V). Since S = P / pf = kW × 1000 / pf, the formula becomes I = (kW × 1000) / (√3 × V × pf). For example, a 5.543 kW load at 400 V and pf 0.8 draws I = 5543 / (1.732 × 400 × 0.8) = 10 A.
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